= Solution
Two integral binary quadratic forms are properly equivalent when one is obtained from the other by a change of variables in $SL_2(\mathbb Z)$. For a negative discriminant $d\equiv0,1\pmod4$, the <class number of a negative discriminant>[class number of binary quadratic forms] $h(d)$ is the number of proper equivalence classes of primitive positive-definite forms of discriminant $d$.
If $f$ properly represents $m$, then $f(r,s)=m$ for coprime $r,s$. Extend $(r,s)^T$ to a matrix in $SL_2(\mathbb Z)$; after the associated variable change, the coefficient of $x^2$ is $m$, so the transformed form is
$$
mx^2+bxy+cy^2.
$$
The converse follows by evaluating this form at $(1,0)$.
Such a form has discriminant $d$ precisely when
$$
b^2-4mc=d,
$$
so it exists exactly when $b^2\equiv d\pmod{4m}$. Thus $m$ is properly represented by some form of discriminant $d$ exactly when $d$ is a square modulo $4m$.
Now let $d=1-4A$. If $n^2+n+A$ is composite for $0\leq n\leq A-2$, it has a prime divisor $p<A$, and
$$
d=(2n+1)^2-4(n^2+n+A)\equiv(2n+1)^2\pmod{4p}.
$$
Conversely, if $d$ is a square modulo $4p$ for a prime $p<A$, choose an odd square root $2n+1$ with $0\leq n<p$. Then $p\mid n^2+n+A$ and $n\leq A-2$, so that value is composite.
Finally, reduction theory says that every nonprincipal positive-definite class of discriminant $1-4A$ has a reduced representative whose leading coefficient has a prime divisor $p<A$ for which $d$ is a square modulo $4p$. Conversely such a prime produces a represented form not equivalent to the principal form $x^2+xy+Ay^2$. Combining this with the preceding equivalence gives
$$
\boxed{h(1-4A)=1\iff n^2+n+A\text{ is prime for }0\leq n\leq A-2}.
$$
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